P6: Reflectionless Coatings and

The Double Discontinuity on a String

Adapted from Joe Rogers, Fall 1996

In optical design, one often wants to put a thin film on a glass surface to minimize the amplitude of a reflected wave (and maximize the transmitted wave). This is the principle, for example, behind reflectionless coatings on eyeglasses. The layer between the glass and the air has a wave velocity intermediate between that of the glass and the air. By making the film thickness an appropriate number of wavelengths of the incoming light, one can arrange for destructive interference between the light reflected off the air-film boundary and off the film-glass boundary.

There is an analog of this optical thin film for a wave on a string. Suppose the left end of a string has a high mass density (and therefore a low wave velocity), and the right end of the string has a low mass density (and a high wave velocity). How can we get a wave to pass from the left to the right wieth as little reflection and as much transmission as possible? We can put a short piece of string between the heavy and the light pieces, which has a mass density and wave velocity intermediate between those of the heavy and light pieces.

(P6a) Single Discontinuity

Run pythag, and select SingJump from the preset menu. On the Y vs. X plot at the top, you should see a thick string on the left and a thin one on the right, with the boundary at x=8. A wiggly pulse is coming in from the left. We place the boundary between the ropes far to the right so that we can better see the reflected pulse. It is partially transmitted and partly reflected. Notice that the Y vs. T plot on the bottom does not measure the signal in the middle of the rope, nor at the boundary between the two segments: it measures the motion at the black dot at x=1. Using the Y vs. T plot, measure and write down the ratio of reflected to incident amplitudes.

Now, hit the Configure button, and note the three mass densities mu1, mu2, and mu3. The rope has density mu1=0.1 kg/m for 0<x<X12, has density mu2=0.05 kg/m for X12<x<X23, and has density mu3=0.025 kg/m for X23<x<L. For this problem, since both steps are at x=8, the discontinuity takes us directly from mu1 to mu3. Using these mass densities, calculate and write down the theoretical ratio of reflected to incident amplitudes you expect. To what percentage is this accurate?

(P6b) Reflectionless Coating.

Now for the real challenge! Using the Configure menu, change X23 to (for example) 8.5. Now we have a third segment to the rope, with the intermediate density mu2. Calculate and record the wavelength of the wave in this intermediate, red section of string. If you like, you can check your calculation by enlarging the red section of the rope and halting pythag when the pulse is inside, but be sure to shrink it again afterward! Calculate and record the smallest length DX of the intermediate density of string that minimizes the reflection for this wave (with angular frequency omega=300 radians/sec). Shift X23 to make the red section of the rope this length, and measure and write down the ratio of reflected to incident maximum amplitudes. Did it substantially decrease the reflection? (If you are insecure about your answer, my answer can be found in the preset RefJump).

(P6c) Further Explorations.

What happens, for the length of red string you used in part (b) to minimize the reflection, if you double the frequency omega to 600 radians/sed? Compare it to the intensity of reflection you got in part (a) for a single step. Explain why they are related.

Set omega back to 300 radians/sec, and see what happens as you vary the length of the red string a small amount. Try making the red string 1.2*DX in length, and 0.8*DX in length, to make sure that the destructive interference condition you derived in (b) really minimizes the reflection. Hint: comparing the size of two pulses is easy if you Copy Graph for Y vs. T after the first run, steal data after the second run, and zoom in on the reflected pulses.

Calculate and record the second- and third-shortest lengths of red string that ought to give destructive interference, in terms of the shortest length DX. Compare the amplitude of the reflected pulses with that of the shortest piece of string. If the pulses were perfect sinusoidal waves with precise wavelengths, then the reflected intensity wouldn't change as the path length difference changes by a multiple of the wavelength. Our pulse is not a perfect sine wave: because the pulse has a FWHM width of about four, it must have a small range of wave-lengths by the uncertainty principle (which we'll discuss soon in class!) The longer the distance between the two reflecting boundaries, the more effect this spread of wavelengths has, and the less destructive the interference. This is one reason why you don't see interference fringes for thick films (like windowpanes)!

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Statistical Mechanics: Entropy, Order Parameters, and Complexity, now available at Oxford University Press (USA, Europe).